Rule de l'Hôpital - Continuous and Differentiable

[mk9898]

Homework Statement

Prove that

$h: [0,\infty) \rightarrow \mathbb R, x \mapsto \begin{cases} x^x, \ \ x>0\\ 1, \ \ x = 0\\ \end{cases}$
is continuous but not differentiable at x = 0.

The Attempt at a Solution

To show continuity, the limit as x approaches 0 from the right must equal to 1. Meaning:

$\lim_{x \ \downarrow \ 0} h(x) = 1$.

$\lim_{x \ \downarrow 0} h(x) = \lim_{x \ \downarrow 0} x^x = 0^0 = 1$ (I'm not too content with this part. Just plugging in 0 seems wrong to me and I would have to maybe use the definition of continuity?)

Differentiable:
$h'(0) = 0$
$\lim_{x \ \downarrow 0} \frac{f(x) - f(0)}{x-0} = \lim_{x \ \downarrow 0} \frac{x^x - 0^0}{x-0} = \lim_{x \ \downarrow 0} \frac{x^x - 1}{x} = \lim_{x \ \downarrow 0} x^{x-1}-1$ does not exists due to 1/0. Therefore h(x) is not differentiable.

 

[Orodruin]

There is no left side. The function is not defined for $x < 0$.

 

[mk9898]

Crap. Right. Fixed it.

 

[mk9898]

Can anyone give me some insight please?

 

[mk9898]

I made an error in my calculations. I think I got it now.

 

[Dick]

$\lim_{x \ \downarrow 0} h(x) = \lim_{x \ \downarrow 0} x^x = 0^0 = 1$ (I'm not too content with this part. Just plugging in 0 seems wrong to me and I would have to maybe use the definition of continuity?)

Differentiable:
$h'(0) = 0$
$\lim_{x \ \downarrow 0} \frac{f(x) - f(0)}{x-0} = \lim_{x \ \downarrow 0} \frac{x^x - 0^0}{x-0} = \lim_{x \ \downarrow 0} \frac{x^x - 1}{x} = \lim_{x \ \downarrow 0} x^{x-1}-1$ does not exists due to 1/0. Therefore h(x) is not differentiable.

I'm not too happy with what you are doing there either. Maybe you should use that $x^x=e^{x \ln x}$ to get some believable arguments.

 

[mk9898]

Thanks for the response. I figured that my last step there is false. After L'Hospital I found the answer.